VJesus12 wrote:A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. If there is only one way to present each letter, how many different codes are possible?

A. 42
B. 210
C. 420
D. 840
E. 5,040

-------ASIDE------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
----ONTO THE QUESTION--------------------

We want to arrange A, A, A, B, B, C, and D
There are 7 letters in total
There are 3 identical A's
There are 2 identical B's
So, the total number of possible arrangements = 7!/[(3!)(2!)
= (7)(6)(5)(4)(3)(2)(1)/(3)(2)(1)(2)(1)
= (7)(6)(5)(4)/(2)(1)
= (7)(6)(5)(2)
= 420

Answer: C

Cheers,
Brent